1.

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200 K to 400 K ? (R= 8.314"kJ"mol^(-1)K^(-1))

Answer»

`234.65"KJ MOL"^(-1)K^(-1)`
`434.65"kJ mol"^(-1)K^(-1)`
`434.65"J mol"^(-1)K^(-1)`
`334.65"kJ mol"^(-1)K^(-1)`

Solution :`T_1=200K,k=k_1`
`T_2=400K , k = k_2=2k_1`
`log ((k_2)/(k_1))=(2.303E_a)/R((T_2-T_1)/(T_1t_2))`
`log((2k_2)/(k_1))=(2.303E_a)/(8.314JK^(-1)mol^(-1))((400K-200K)/(200Kxx400K))`
`E_a=(0.3010xx8.314"J mol"^(-1)xx200xx400)/(2.303xx200)`
`E_a=434.65"J mol"^(-1)`


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